Need some urgent help, this is from an exam question. The question requires
us to work out the expression for Io/Vi

This is what i have tried so far,
oamp 1 e+ = (Vi/2)
using e+ = e- The base of transistor is now Vi/2,
i assumed Ve = Vi/2 as well (or should it be Vi/2 - 0.7)
Now ie = Vi/2/R3 and ie = ic thus Vc = Vi/2 + 0.2 (say saturated)

Now e+ of opamp2 is (Vi/2 + 0.2) say Vi/2 again thus e- = Vi/2
Thus the voltage drop across the resistors is V-Vi/2.
Not sure after this, i understand some aspects of the PMOS, when VGS is < Vt then current flows.

Do you need to concern yourself with the detail of the MOSFET characteristics, or indeed those of the transistor? For this circuit to be functioning in a predictable way, it seems likely that we should assume that both the amplifiers will generate suitable output voltages so that their differential input voltages become quite small.

Now e+ of opamp2 is (Vi/2 + 0.2) say Vi/2 again thus e- = Vi/2
Thus the voltage drop across the resistors is V-Vi/2.
Not sure after this, i understand some aspects of the PMOS, when VGS is < Vt then current flows.

Click to expand...

What could be important to state here is:
* The current over the transistor is proportional to Vi because Ve = Vi/2 and R3 is constant

If V is also constant, the same for Voltage_R2, and therefore Ve+(2) is function of Vi, so for Ve-(2) (assumed negative feedback through Mosfet).